JEE Main 2026 · 6 Apr · Shift 1, Physics question 46

Polarity (as drawn). Call the junction after the 3 Ω resistor X and the right-hand wire Y, and let $V=V_X-V_Y$.

• Main branch: $V_X=V_Y+2-3I$.
• 6 Ω branch: $(V-3)/6$ from X to Y.
• 4 Ω branch: $V/4$.

KCL.
$$\frac{V-3}6+\frac V4=\frac{2-V}3\ \Rightarrow\ 9V=14,\qquad V=\frac{14}9.$$

Current in 6 Ω.
$$\left|\frac{14/9-3}6\right|=\frac{13}{54}\text{ A}.$$

Heat in 100 s.
$$\left(\frac{13}{54}\right)^2(6)(100)=34.77\text{ J}=\frac{3477}{100}\text{ J}.$$

Answer: 3477